2
$\begingroup$

There has been a number of helpful threads clarifying the essence of the notion of the opposite category (particularly here and here). However I am still confused with taking the opposite of the category where sets of morphisms between two objectes are themselves objects of the category (e.g. Set or R-mod).

Here is the source of confusion:

Let us take the category $\mathcal{C}$ to be R-mod for commutative $R$ with identity.

Now for $R$-modules $M,N$, the set $\operatorname{Hom}_{R-mod}(M,N)$ is again an $R$-module and for any morphism $M_1 \xrightarrow{\phi} M_2$ we have the induced maps $$(*) \ \ \ \ \ \ \operatorname{Hom}_{R-mod} (M_1, N) \xleftarrow{\phi^*} \operatorname{Hom}_{R-mod} (M_2, N),$$ $$\ \ \ \ \ \ \ \ \ \ \ \operatorname{Hom}_{R-mod} (P, M_1) \xrightarrow{\phi_*} \operatorname{Hom}_{R-mod} (P, M_2),$$ which are $R$-linear, so that they are morphisms in $\mathcal{C}$ = R-mod.

Therefore in $\mathcal{C}^{op}$ = R-mod$^{op}$ we obtain a morphisms "opposite" to $\phi^*$ and $\phi_*$:

$$(**) \ \ \ \ \ \ \operatorname{Hom}_{R-mod^{op}}(N^{op}, M_{1}^{op}) \xrightarrow{({\phi^{*}})^{op}}\operatorname{Hom}_{R-mod^{op}}(N^{op}, M_{2}^{op}),$$

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname{Hom}_{R-mod^{op}}(M_{1}^{op}, P^{op}) \xleftarrow{({\phi_*})^{op}}\operatorname{Hom}_{R-mod^{op}}(M_{2}^{op}, P^{op}),$$

where I am using the identification $(\operatorname{Hom}_{R-mod}(X, Y))^{op} = \operatorname{Hom}_{Y^{op}}(Y^{op}, X^{op})$.

On the other hand, the opposite to $\phi$, $M_1^{op} \xleftarrow{\phi^{op}} M_2^{op}$, induces maps in the inverse directions to $(**)$:

$$(***) \ \ \ \ \ \ \operatorname{Hom}_{R-mod^{op}}(N^{op}, M_{1}^{op}) \xleftarrow{{(\phi^{op}})_*}\operatorname{Hom}_{R-mod^{op}}(N^{op}, M_{2}^{op}),$$

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname{Hom}_{R-mod^{op}}(M_{1}^{op}, P^{op}) \xrightarrow{{(\phi^{op}})^*}\operatorname{Hom}_{R-mod^{op}}(M_{2}^{op}, P^{op}),$$

  1. Do I understand it correctly, that $(***)$ are just set-maps and not morphisms in R-mod$^{op}$, since there is no reason to believe that there are corresponding morphisms $$\operatorname{Hom}_{R-mod} (M_1, N) \to \operatorname{Hom}_{R-mod} (M_2, N)$$ $$\operatorname{Hom}_{R-mod} (P,M_1) \leftarrow \operatorname{Hom}_{R-mod} (P, M_2)$$ in R-mod (these directions are unnatural)?
  2. If 1 is true, how then to make sense that in R-mod morphisms between objects induce morphisms between Hom's (see $(*)$), but in R-mod$^{op}$ we merely get set-theoretic maps $(***)$ which are not morphisms in $\mathcal{C}^{op}$? Doesn't that seem unnatural, since in my mind the opposite category should mimic the properties of $\mathcal{C}$ very closely.
  3. Closely related to 2, it seems that R-mod comes equipped with just a pair of maps $(*)$, whereas R-mod$^{op}$ comes equipped with two pairs, $(**)$ and $(***)$, which also seems strange.

PS. My goal is to understand the fact that contravariant $h_N=\operatorname{Hom}_{R-mod} ( - , N)$ is right-adjoint to itself meaning that we view $h_N :$ R-mod $\to $ R-mod$^{op}$ as right adjoint to $h_N^{op}:$ R-mod$^{op}$ $\to$ R-mod as explained in the algebra textbook of Paolo Aluffi. On the level that we "just have to reverse all arrows to account for contravariant nature of $h_N$" I understand his argument, but rigorous understanding of what is going on in the opposite categories evades me.

$\endgroup$
1
$\begingroup$

The proper way to understand this is not by fixing a single $\phi$. Let $C$ be the category of $R$-modules and $C^*$ its opposite. You have a composition map of modules, $C(N,P)\otimes C(M,N)\to C(M,P), f\otimes g\mapsto f\circ g$. In the opposite category, we have $C^*(P,N)=C(N,P)$ (where we don't distinguish between a module viewed in $C$ versus $C^*$.) So the above composition becomes a map (in $C$!) $C^*(P,N)\otimes C^*(N,M)\to C^*(P,M)$. Applying the symmetry isomorphism for the tensor product of modules, this is essentially the same as a map (still in $C$!) $C^*(N,M)\otimes C^*(P,N)\to C^*(P,M)$, which is just the composition map for $C^*$. Every hom-object and map thereof I've mentioned is an $R$-module, so we see $C^*$ still has homs and composition maps from $C$.

It's also perfectly possible to view $\otimes$ as a functor $C^*\times C^*\to C^*$. But that doesn't mean that the same composition map makes $C^*$'s composition be a map of $C^*$-objects: it goes in the wrong direction for that! So in (**), it wasn't meaningful to make the morphisms of hom-modules lie in $C^*$: you should leave them in $C$. This is not so unfamiliar: after all, the opposite of a category is a category, not something with homs from complete atomic Boolean algebras (that is, from the opposite of sets!)

In (***), you're missing something entirely. No such morphisms exist, as you almost prove right away: there are neither module maps nor set functions $C(M_1,N)\to C(M_2,N)$ induced by $\phi$. What could they possibly be?

The situation we're summarizing is leading toward the general theory of "enriched categories", which are splice categories but with homs forming something more interesting than a set. It's always the case that, when defined, the opposite of an enriched category is enriched over the same category, not a new one.

Anyway, the adjunction you're trying to describe has nothing to do with any of these issues. Let $F(M)=C(M,N)$ as an $R$-module, Then $C(P,F(M))=C(M,F(P))$, which by definition is $C^*(F(P),M)$. There's really no need for deep reflection on opposite categories here, although it might well be worthwhile in its own right.

$\endgroup$
  • $\begingroup$ thank you for your answer. I don't understand your comment about (***): isn't it true that for any category a given $M_1 \xrightarrow{\phi} M_2$ induces set-theoretic maps on sets of morphisms (which form sets by axioms of a category) by pulling-back (precomposing - $\alpha \mapsto \alpha \circ \phi$) or post-composing ($\beta \mapsto \phi \circ \beta$)? How is $\mathcal{C}^{op}$ different in this regard? $\endgroup$ – Bananeen Nov 14 '16 at 4:12
  • $\begingroup$ Oh, I misread what you wrote. The arrows you drew are just maps in $C$, the same ones I described. The thing you draw in question 1 is a map of $C^*$: you're flipping too much. $\endgroup$ – Kevin Carlson Nov 14 '16 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.